Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, par[100005];
vector<int> adj[100005];
vector<int> cit[100005];
set<int> an;
struct node {
  int v = 0;
  node* left = NULL;
  node* right = NULL;
};
node* root[100005];
void build(node* t, int l, int r) {
  if (l == r) {
    t->left = NULL;
    t->right = NULL;
    t->v = 0;
    return;
  }
  int mid = (l + r) >> 1;
  node* a = new node;
  a->left = NULL;
  a->right = NULL;
  a->v = 0;
  t->left = a;
  node* b = new node;
  b->left = NULL;
  b->right = NULL;
  b->v = 0;
  t->right = b;
  build(t->left, l, mid);
  build(t->right, mid + 1, r);
  t->v = t->left->v + t->right->v;
}
void upd(node* t, int l, int s, int e) {
  if (s == e) {
    t->v = 1;
    t->left = NULL;
    t->right = NULL;
    return;
  }
  int mid = (s + e) >> 1;
  if (s <= l && mid >= l) {
    node* a = new node;
    a->left = t->left->left;
    a->right = t->left->right;
    a->v = t->left->v;
    t->left = a;
    upd(t->left, l, s, mid);
  } else if (mid + 1 <= l && e >= l) {
    node* a = new node;
    a->left = t->right->left;
    a->right = t->right->right;
    a->v = t->right->v;
    t->right = a;
    upd(t->right, l, mid + 1, e);
  }
  t->v = t->left->v + t->right->v;
}
int tree[2000000][2];
int tour[500005];
int level[500005];
int ind[100005];
int inde = 1;
pair<int, int> query(int l, int r, int s, int e, int i) {
  if (s > r || e < l) return (make_pair(1000000000, -1));
  if (s >= l && e <= r) return (make_pair(tree[i][0], tree[i][1]));
  int mid = (s + e) / 2;
  pair<int, int> a, b;
  a = query(l, r, s, mid, 2 * i);
  b = query(l, r, mid + 1, e, 2 * i + 1);
  return min(a, b);
}
void build1(int i, int l, int r) {
  if (l == r) {
    tree[i][0] = level[l];
    tree[i][1] = l;
    return;
  }
  int mid = (l + r) / 2;
  build1(2 * i, l, mid);
  build1(2 * i + 1, mid + 1, r);
  if (tree[2 * i][0] < tree[2 * i + 1][0]) {
    tree[i][0] = (tree[2 * i][0]);
    tree[i][1] = tree[2 * i][1];
  } else {
    tree[i][0] = (tree[2 * i + 1][0]);
    tree[i][1] = tree[2 * i + 1][1];
  }
}
int lca(int x, int y) {
  x = ind[x];
  y = ind[y];
  if (x > y) swap(x, y);
  return tour[query(x, y, 1, inde - 1, 1).second];
}
void dfs(int i, int l, int p) {
  par[i] = p;
  level[inde] = l;
  ind[i] = inde;
  tour[inde] = i;
  inde++;
  root[i] = new node;
  root[i]->left = root[p]->left;
  root[i]->right = root[p]->right;
  root[i]->v = root[p]->v;
  for (auto j : cit[i]) upd(root[i], j, 1, m);
  for (int j = 0; j < adj[i].size(); j++) {
    if (adj[i][j] != p) {
      dfs(adj[i][j], l + 1, i);
      tour[inde] = i;
      level[inde] = l;
      inde++;
    }
  }
}
void Q(node* a, node* b, int s, int e, int k) {
  if (k == 0 || b->v - a->v == 0) return;
  if (s == e) {
    an.insert(s);
    return;
  }
  int cnt = b->left->v - a->left->v;
  int mid = (s + e) >> 1;
  Q(a->left, b->left, s, mid, k);
  if (cnt < k) Q(a->right, b->right, mid + 1, e, k - cnt);
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &q);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    cit[x].push_back(i);
  }
  root[0] = new node;
  build(root[0], 1, m);
  dfs(1, 0, 0);
  build1(1, 1, inde - 1);
  for (int i = 1; i <= q; i++) {
    an.clear();
    int x, y, a;
    scanf("%d", &x);
    scanf("%d", &y);
    scanf("%d", &a);
    int l = par[lca(x, y)];
    int k = min(root[x]->v - root[l]->v, 2 * a);
    Q(root[l], root[x], 1, m, k);
    k = min(root[y]->v - root[l]->v, 2 * a);
    Q(root[l], root[y], 1, m, k);
    int s = an.size();
    int cnt = min(a, s);
    printf("%d ", cnt);
    for (int j = 1; j <= cnt; j++) {
      int h = *an.begin();
      an.erase(an.begin());
      printf("%d ", h);
    }
    printf("\n");
  }
  return 0;
}